Question 1

The sum of two numbers is $528$ and their H.C.F. is $33$. The number of pairs of numbers satisfying the above conditions is

Accepted Answer

Concept & Formula: Let the two numbers be 33a and 33b, where a and b are strictly co-prime integers (H.C.F.(a, b) = 1). Using the sum condition: 33a + 33b = Sum The number of valid co-prime pairs (a, b) determines the total number of pairs of numbers that satisfy the given conditions. Step-by-Step Solution: • Given: Sum = 528, H.C.F. = 33. • Step 1: Represent the numbers as 33a and 33b. • Step 2: Formulate the equation: 33a + 33b = 528 33(a + b) = 528 • Step 3: Isolate (a + b): a + b = (528/33) = 16 • Step 4: List all positive integer pairs (a, b) such that a + b = 16, and filter for co-prime pairs: • (1, 15) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (2, 14) \rightarrow Both even (H.C.F. = 2). (Invalid) • (3, 13) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (4, 12) \rightarrow Share factor 4. (Invalid) • (5, 11) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (6, 10) \rightarrow Both even (H.C.F. = 2). (Invalid) • (7, 9) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (8, 8) \rightarrow Share factor 8. (Invalid) • Step 5: Count the valid pairs: (1, 15), (3, 13), (5, 11), and (7, 9). There are exactly 4 pairs. Exam Strategy & Shortcut: Coprimality Count: Once you find a + b = 16, you can find the number of valid pairs by calculating how many numbers less than 16 are co-prime to 16 (Euler's totient function \phi(16)) and dividing by 2 (since order doesn't matter for pairs). Since 16 = 2^4, the numbers not co-prime to it are all even numbers. The odd numbers less than 16 are 1, 3, 5, 7, 9, 11, 13, 15 (8 numbers). Pairs = 8 / 2 = 4. This avoids listing and checking every single pair manually. Common Pitfall: A very common mistake is counting all possible additive combinations that sum to 16 (which would give 8 pairs, leading to option c). Always remember to filter out any pairs that share a common factor, as their inclusion would change the H.C.F. from 33 to a higher multiple. Final Answer: Therefore, the correct answer is 4.

Question 2

The L.C.M. and ratio of four numbers are 630 and 2 : 3 : 5 : 7 respectively. The difference between the greatest and least numbers is

Accepted Answer

Concept & Logic: The L.C.M. of numbers given in a ratio where the ratio terms are mutually coprime is simply the product of those ratio parts multiplied by their Highest Common Factor (H.C.F.). L.C.M. = H.C.F. × (Product of coprime ratio parts) Step-by-Step Solution: • Let the common factor (H.C.F.) be x. The four numbers can be represented as 2x, 3x, 5x, and 7x. • Because the numbers 2, 3, 5, and 7 are prime to each other (coprime), their L.C.M. is the product of these ratios and x. • L.C.M. = 2 × 3 × 5 × 7 × x = 210x. • We are given that the L.C.M. is 630. • Equating the two: 210x = 630. • Solving for x: x = 630 / 210 = 3. • The greatest number is 7x = 7 × 3 = 21. • The least number is 2x = 2 × 3 = 6. • The difference between the greatest and least numbers is 21 - 6 = 15. Exam Strategy & Shortcut: Look at the ratio: the difference between the greatest part (7) and the least part (2) is 5 parts. This means the final answer MUST be a multiple of 5. Scanning the given options (6, 14, 15, 21), 15 is the only multiple of 5. You can confidently select it without calculating the L.C.M. or solving for x. Common Pitfall: A frequent mistake is solving for x = 3 and then mistakenly choosing option (a) which is 6, or option (d) which is 21, by forgetting that the question specifically asks for the *difference* between the two extreme numbers, not just one of the numbers. Final Answer: Therefore, the correct answer is 15.

Question 3

The H.C.F. and L.C.M. of two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4, then the larger of the two numbers is

Accepted Answer

Concept & Formula: When two numbers are given in a ratio a : b, the actual numbers can be represented as ax and bx, where the common multiplier x is the Highest Common Factor (H.C.F.) of the two numbers. Numbers = (Ratio Part 1 × H.C.F.) and (Ratio Part 2 × H.C.F.) Step-by-Step Solution: • Given: * H.C.F. = 21 • L.C.M. = 84 • Ratio of numbers = 1 : 4 • Let the two numbers be 1x and 4x. • By definition, the common factor x is the H.C.F. Therefore, x = 21. • The two numbers are: • First number = 1 × 21 = 21 • Second number = 4 × 21 = 84 • The question asks for the larger of the two numbers. • Comparing 21 and 84, the larger number is 84. Exam Strategy & Shortcut: You don't even need the L.C.M. to solve this specific question. The problem explicitly gives the ratio (1:4) and the H.C.F. (21). The larger number corresponds to the larger ratio part (4). Simply multiply the larger ratio part by the H.C.F.: 4 × 21 = 84. You can solve this in 3 seconds mentally. Common Pitfall: A common trap is assuming you need to use all the provided numbers (like the L.C.M. = 84) in a complex formula. While you *could* use the formula Product = H.C.F. × L.C.M., it unnecessarily complicates a straightforward ratio multiplication. Don't over-engineer the solution. Final Answer: Therefore, the correct answer is 84.

Question 4

The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?

Accepted Answer

Concept & Logic: The core rule relating these two metrics is that the Highest Common Factor (H.C.F.) of any given set of numbers must be a perfect factor (divisor) of their Lowest Common Multiple (L.C.M.). If an L.C.M. value is not a multiple of the H.C.F., it is an invalid L.C.M. L.C.M. = k × H.C.F. (where k is an integer) Step-by-Step Solution: • Given: H.C.F. = 8. • We are looking for an option that is NOT a multiple of 8 (i.e., 8 does not divide it without a remainder). • Let's evaluate the choices: • Option (a) 24: 24 / 8 = 3. (Valid L.C.M. candidate). • Option (b) 48: 48 / 8 = 6. (Valid L.C.M. candidate). • Option (c) 56: 56 / 8 = 7. (Valid L.C.M. candidate). • Option (d) 60: 60 / 8 = 7.5. (Invalid candidate; leaves a remainder of 4). • Because 60 is not perfectly divisible by 8, it can never represent the L.C.M. for this pair of numbers. Exam Strategy & Shortcut: This question strictly tests your knowledge of multiplication tables. You should know the multiples of 8 by heart: 8, 16, 24, 32, 40, 48, 56, 64. By rapidly scanning the options against this list, 60 immediately stands out as the odd one out. Common Pitfall: Similar to other negative-constraint questions, students rushing under time pressure might read "can never be" as "can be" and erroneously select 24 simply because it is the first valid multiple they spot. Always double-check what the question is specifically asking you to identify. Final Answer: Therefore, the correct answer is 60.

Question 5

If the L.C.M. of three numbers is 9570, then their H.C.F. can be

Accepted Answer

Concept & Formula: The fundamental rule relating Highest Common Factor (H.C.F.) and Lowest Common Multiple (L.C.M.) of any set of numbers is that the H.C.F. must always be a perfect divisor of the L.C.M. L.C.M. = k × H.C.F. where k is a positive integer. Step-by-Step Solution: • Given: L.C.M. = 9570. • Calculation / Deduction: We must test which of the given options perfectly divides 9570 without leaving a remainder. • Testing 11: 9570 / 11 = 870 (Exact integer). • Testing 12: 9570 / 12 = 797.5 (Decimal). • Testing 19: 9570 / 19 \approx 503.68 (Decimal). • Testing 21: 9570 / 21 \approx 455.71 (Decimal). Exam Strategy & Shortcut: Use divisibility rules to immediately spot the answer without full long division. For 11, check the alternating sum of digits: (9 + 7) - (5 + 0) = 16 - 5 = 11. Since 11 is perfectly divisible by 11, the number 9570 is as well. You can confidently select option (a) and move on in seconds. Common Pitfall: Students often think they need to find the specific three numbers first to determine the H.C.F. This is impossible with the given data and leads to a dead end. Always rely on the primary relationship rule between L.C.M. and H.C.F. Final Answer: Therefore, the correct answer is 11.

Question 6

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is

Accepted Answer

Concept & Logic: When two numbers have an H.C.F. of H, the numbers can be represented as Hx and Hy, where x and y are co-prime factors. The L.C.M. of these numbers is given by: L.C.M. = H × x × y The problem explicitly provides H and the other two factors (x and y). Step-by-Step Solution: • Given: H.C.F. = 23, and the other two factors of the L.C.M. are 13 and 14. • Calculation / Deduction: Let the two numbers be N_1 and N_2. • Based on the concept, the numbers are 23 × 13 and 23 × 14. • The problem asks for the larger of the two numbers. • Larger number = 23 × 14. • 23 × 14 = 322. Exam Strategy & Shortcut: You don't need to calculate both numbers. Simply identify the larger factor (14) and multiply it by the H.C.F. (23). To quickly calculate 23 × 14, use the unit digit trick: 3 × 4 = 12, so the answer must end in 2. Looking at the options, only (c) 322 ends in 2, making it an instant pick. Common Pitfall: A common mistake is multiplying all three numbers together (23 × 13 × 14) assuming the question asks for the L.C.M., or mistakenly assuming 13 and 14 are the numbers themselves rather than the co-prime factors. Read the prompt carefully. Final Answer: Therefore, the correct answer is 322.

Question 7

Two numbers, both greater than 29, have H.C.F. 29 and L.C.M. 4147. The sum of the numbers is

Accepted Answer

Concept & Formula: When given the H.C.F. of two numbers, we can define the numbers as Hx and Hy, where H is the H.C.F., and x and y are co-prime integers. L.C.M. = H × x × y Step-by-Step Solution: • Given: H.C.F. (H) = 29, L.C.M. = 4147. Both numbers are greater than 29. • Calculation / Deduction: Let the numbers be 29x and 29y. • Using the formula: 29 × x × y = 4147. • Solve for xy: xy = 4147 \div 29 = 143. • We must find co-prime factors of 143. By testing small primes, we see 143 = 11 × 13. • So, x = 11 and y = 13. • The numbers are 29 × 11 = 319 and 29 × 13 = 377. • Both are strictly greater than 29, satisfying the given condition. • Find the sum: 319 + 377 = 696. Exam Strategy & Shortcut: Use unit digit estimation to verify calculations quickly. For the sum 29(11 + 13) = 29 × 24. Look at the unit digits: 9 × 4 = 36. The final answer must end in a 6. This eliminates option (b). Between 666 and 696, since 30 × 24 = 720, the answer 29 × 24 must be just slightly less than 720 (720 - 24 = 696). Common Pitfall: Ignoring the condition "both greater than 29". If a student factored 143 as 1 × 143, the numbers would be 29 × 1 = 29 and 29 × 143 = 4147. While their H.C.F. is 29 and L.C.M. is 4147, one number is exactly 29, violating the "greater than 29" rule. Always test the co-prime pair that are closer in value. Final Answer: Therefore, the correct answer is 696.

Question 8

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Accepted Answer

Concept & Formula: When asked to find the greatest number that divides x, y, and z leaving the same remainder in each case, the required number is given by taking the HCF of their absolute differences: Required Number = HCF of (|x - y|, |y - z|, |z - x|) Step-by-Step Solution: • Given: Numbers are 43, 91, and 183. • Step 1: Find the absolute differences between the given numbers pairwise. 91 - 43 = 48 183 - 91 = 92 183 - 43 = 140 • Step 2: Calculate the HCF of 48, 92, and 140. • Let's break them down: 48 = 4 × 12 92 = 4 × 23 140 = 4 × 35 • The numbers 12, 23, and 35 have no common prime factors. • Therefore, the highest common factor among these differences is 4. Exam Strategy & Shortcut: Option Elimination is vastly faster here! Instead of doing the difference and HCF calculations, just plug the options into the original numbers to check for identical remainders. Dividing by 4 gives a remainder of 3 in all cases (43 = 40+3, 91 = 88+3, 183 = 180+3). You can verify this in under 10 seconds. Common Pitfall: A standard trap is trying to find the HCF of 43, 91, and 183 directly. This won't work because they don't share a common factor, and it completely ignores the "same remainder" constraint. Always remember to take the differences first for this specific question type. Final Answer: Therefore, the correct answer is 4.

Question 9

The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case is

Accepted Answer

Concept & Formula: When asked to find the greatest number that divides given numbers x, y, and z leaving the same specific remainder R in each case, the required number is the Highest Common Factor (HCF) of the numbers obtained by subtracting the remainder from the original numbers. Required Number = HCF of (x - R), (y - R), and (z - R) Step-by-Step Solution: Given: • Numbers: 1356, 1868, and 2764 • Common Remainder: 12 Calculation: • Step 1: Subtract the common remainder from each of the given numbers. • 1356 - 12 = 1344 • 1868 - 12 = 1856 • 2764 - 12 = 2752 • Step 2: Find the HCF of the resulting numbers (1344, 1856, 2752). • Step 3: Use the difference method to simplify finding the HCF. Find the differences between adjacent numbers. • 1856 - 1344 = 512 • 2752 - 1856 = 896 • Step 4: Find the HCF of these differences (512 and 896). • 896 - 512 = 384 • 512 - 384 = 128 • 384 \div 128 = 3 (Perfectly divisible) • So, the HCF of the differences is 128. The final HCF must be a factor of 128. • Step 5: Check the options or factors of 128. The given numbers (1344, 1856, 2752) are not perfectly divisible by 128 (e.g., 1344 \div 128 = 10.5). The next largest factor of 128 is 64. • 1344 \div 64 = 21 • 1856 \div 64 = 29 • 2752 \div 64 = 43 • Since 21, 29, and 43 are co-prime, the HCF is definitively 64. Exam Strategy & Shortcut: Instead of manually calculating the HCF of large numbers, rely on the options and divisibility rules. After finding the adjusted numbers (1344, 1856, 2752), perform a quick scan of the choices. Since 1344 = 1300 + 44, which isn't obviously divisible by large awkward numbers like 124 or 156, test the cleanest option (64) first. It perfectly divides all three, immediately yielding the answer. Common Pitfall: A major mistake is calculating the HCF of the original numbers (1356, 1868, 2764) without subtracting the remainder first. Always remember that the remainder represents the "extra" amount preventing perfect divisibility. It must be stripped away before HCF logic applies. Final Answer: Therefore, the correct answer is 64.

Question 10

Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?

Accepted Answer

Concept & Formula: When a question asks for the greatest number that divides given numbers x and y leaving specific remainders a and b respectively, the required number is the Highest Common Factor (HCF) of the numbers obtained by subtracting the remainders from the original numbers. Required Number = HCF of (x - a) and (y - b) Step-by-Step Solution: Given: • Numbers: 3026 and 5053 • Respective Remainders: 11 and 13 Calculation: • Step 1: Subtract the respective remainders from the given numbers. • 3026 - 11 = 3015 • 5053 - 13 = 5040 • Step 2: Find the HCF of the resulting numbers (3015 and 5040). • Step 3: Prime factorize or use divisibility rules. Let's look at the numbers. • 3015 ends in 5, so it's divisible by 5. The sum of digits (3+0+1+5=9) is divisible by 9. Thus, it is divisible by 45. • 3015 \div 45 = 67. Since 67 is a prime number, the prime factorization of 3015 is 45 × 67. • Step 4: Check if the other number, 5040, is divisible by these factors. • 5040 is clearly divisible by 9 (sum is 9) and by 5 (ends in 0), so it is also divisible by 45. • 5040 \div 45 = 112. • Step 5: The factors of the two numbers are (45 × 67) and (45 × 112). Since 67 and 112 share no common factors, the HCF is 45. Exam Strategy & Shortcut: Leverage the options and basic divisibility rules. We need the HCF of 3015 and 5040. Since 3015 ends in an odd digit (5), it cannot be divided by an even number. This immediately eliminates options (b) 30 and (d) 60. Now check the remaining options: 15 and 45. Apply the divisibility rule for 9 (sum of digits). The sum of digits for both 3015 and 5040 is 9, meaning both are divisible by 9. Since 15 is not a multiple of 9 but 45 is, the HCF must be 45. Common Pitfall: Students sometimes reverse the remainders (subtracting 13 from 3026 and 11 from 5053). Always ensure you map the remainders strictly in the "respective" order given in the problem statement. Final Answer: Therefore, the correct answer is 45.

Question 11

The smallest fraction, which each of 6/7, 5/14, 10/21 will divide exactly is

Accepted Answer

Concept & Formula: When a question asks for the smallest number or fraction *which is exactly divided by* other fractions, it is asking for the Least Common Multiple (LCM) of those fractions. The formula for finding the LCM of fractions is: LCM of Fractions = \frac{LCM of Numerators}{HCF of Denominators} Step-by-Step Solution: Given fractions: • (6/7), (5/14), (10/21) Calculation: • Step 1: Identify the numerators and denominators. • Numerators: 6, 5, 10 • Denominators: 7, 14, 21 • Step 2: Find the LCM of the numerators (6, 5, 10). • Multiples of 10: 10, 20, 30... • 30 is perfectly divisible by both 6 and 5. • LCM(6, 5, 10) = 30 • Step 3: Find the HCF of the denominators (7, 14, 21). • All numbers are multiples of 7. • HCF(7, 14, 21) = 7 • Step 4: Substitute the values back into the fraction formula. LCM = (30/7) Exam Strategy & Shortcut: Memorize the fraction rules perfectly: "LCM = LCM of top / HCF of bottom". The moment you look at the numerators (6, 5, 10), your mind should instantly jump to 30. Looking at the denominators (7, 14, 21), 7 stands out as the highest common factor. This gives you (30/7) within 5 seconds, allowing you to instantly select option (a). Common Pitfall: A common point of confusion is mixing up the fraction rules and calculating \frac{HCF of Numerators}{LCM of Denominators} instead. That formula finds the *greatest common divisor* (HCF) of fractions, not the smallest common multiple (LCM). Final Answer: Therefore, the correct answer is 30/7.

Question 12

What is the greatest number of $3$ digits which when divided by $6$, $9$ and $12$ leaves a remainder of $3$ in each case?

Accepted Answer

Concept & Logic: To find the greatest n-digit number that leaves a remainder R when divided by a set of numbers, you must follow a three-step process: 1. Find the LCM of the divisors. 2. Find the greatest n-digit multiple of that LCM. 3. Add the required remainder R to that multiple. The general formula is: Required Number = (Greatest n-digit Multiple of LCM) + R Step-by-Step Solution: • Given: Divisors are 6, 9, and 12. Remainder R = 3. We need the greatest 3-digit number. • Find the LCM: 6 = 2 × 3 9 = 3^2 12 = 2^2 × 3 LCM = 2^2 × 3^2 = 4 × 9 = 36 • Find the greatest 3-digit multiple of the LCM: The greatest general 3-digit number is 999. Let's divide 999 by our LCM (36) to find the closest exact multiple. 999 \div 36 = 27 with a remainder of 27. To make 999 perfectly divisible by 36, subtract this remainder: Greatest 3-digit multiple = 999 - 27 = 972 (972 is exactly 36 × 27). • Add the required remainder: The question specifies a remainder of 3 in each case. Required Number = 972 + 3 = 975 Exam Strategy & Shortcut: Divisibility by 9 Reverse-Engineering: The final number leaves a remainder of 3 when divided by 9. Therefore, if we subtract 3 from the correct option, it MUST be perfectly divisible by 9. (Sum of digits must equal a multiple of 9). Let's test: (a) 903 - 3 = 900 \rightarrow Sum = 9 (Possible) (b) 939 - 3 = 936 \rightarrow Sum = 18 (Possible) (c) 975 - 3 = 972 \rightarrow Sum = 18 (Possible) (d) 996 - 3 = 993 \rightarrow Sum = 21 (Eliminated) Now use divisibility by 12 (which means divisible by 4). The number (Option - 3) must be divisible by 4 (last two digits divisible by 4): (a) 900: Last two digits 00 (Divisible by 4). (b) 936: Last two digits 36 (Divisible by 4). (c) 972: Last two digits 72 (Divisible by 4). Since the question asks for the greatest 3-digit number, we simply pick the largest valid option from our list, which is 975. Common Pitfall: The most frequent error is calculating the LCM (36), adding the remainder to get 39, and then trying to find multiples of 39 up to 999. This is mathematically incorrect. You must find the multiple of the LCM first (972) and *then* add the remainder (3) at the very end. Final Answer: Therefore, the correct answer is 975.

Question 13

Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

Accepted Answer

Concept & Formula: To find a number that leaves a remainder r when divided by multiple divisors a, b, c, we use the Least Common Multiple (LCM): \Number = \LCM(a, b, c) \× k + r where k is a positive integer chosen such that the condition for the secondary divisor is fully satisfied. Step-by-Step Solution: Given: * Divisors: 16, 18, 20, 25 • Remainder in each case: 4 • Complete divisibility by: 7 Calculation: 1. Find the LCM of 16, 18, 20, and 25: • 16 = 2^4 • 18 = 2 \× 3^2 • 20 = 2^2 \× 5 • 25 = 5^2 • \LCM = 2^4 \× 3^2 \× 5^2 = 16 \× 9 \× 25 = 3600 2. Express the required number in the standard form: \Number = 3600k + 4 3. Test values of k such that (3600k + 4) is perfectly divisible by 7: • Simplify 3600 \pmod 7: 3600 = 7 \× 514 + 2 • So, the expression becomes: 2k + 4 • For k = 1 \implies 2(1) + 4 = 6 (not divisible by 7) • For k = 2 \implies 2(2) + 4 = 8 (not divisible by 7) • For k = 3 \implies 2(3) + 4 = 10 (not divisible by 7) • For k = 4 \implies 2(4) + 4 = 12 (not divisible by 7) • For k = 5 \implies 2(5) + 4 = 14 (divisible by 7) 4. Substitute k = 5 into the original expression: \Number = 3600 \× 5 + 4 = 18000 + 4 = 18004 Exam Strategy & Shortcut: Option Elimination: Check the options for divisibility by 7 directly. • 17004: 17004 \div 7 = 2429.14 (No) • 18000: Not possible since it leaves a remainder of 0 when divided by 20 and 25, not 4. • 18002: 18002 \div 7 = 2571.71 (No) • 18004: 18004 \div 7 = 2572 (Yes) Only 18004 satisfies the divisibility condition for 7. Common Pitfall: Students often forget to find the specific value of k and mark a value like 3600 + 4 = 3604 if it is in the options, ignoring the divisibility test for the number 7. Always verify the secondary divisibility condition. Final Answer: Therefore, the correct answer is 18004.

Question 14

An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The next time, when they would beep together at the earliest is

Accepted Answer

Concept & Formula: Simultaneous events occurring at different regular intervals repeat together at time durations that are common multiples of their individual intervals. The earliest next occurrence is found using the LCM: \Time Interval = \LCM(\Interval_1, \Interval_2) Step-by-Step Solution: Given: * Interval of first device = 60 seconds • Interval of second device = 62 seconds • Initial simultaneous beep time = 10:00 a.m. Calculation: 1. Find the LCM of 60 and 62: • 60 = 2^2 \× 3 \× 5 • 62 = 2 \× 31 • \LCM = 2^2 \× 3 \× 5 \× 31 = 60 \× 31 = 1860 \ seconds 2. Convert the total seconds into minutes: \Minutes = \(1860/60) = 31 \ minutes 3. Add this interval to the initial starting time: \Next Time = 10:00 \ a.m. + 31 \ minutes = 10:31 \ a.m. Exam Strategy & Shortcut: Instead of computing the large multiplication fully, note that: \LCM(60, 62) = 60 \× \(62/\HCF)(60, 62) = 60 \× \(62/2) = 60 \× 31 \ seconds Dividing by 60 to get minutes immediately cancels out the 60, leaving exactly 31 minutes. Add 31 minutes directly to 10 a.m. to get 10:31 a.m. Common Pitfall: A frequent error is forgetting to convert the calculated LCM from seconds to minutes, or making a manual error during the multiplication step (60 \× 31). Final Answer: Therefore, the correct answer is 10.31 a.m.

Question 15

Four different electronic devices make a beep after every 30 minutes, 1 hour, $1\frac{1}{2}$ hour and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at

Accepted Answer

Concept & Formula: To find when multiple periodic events will synchronize again, we calculate the Least Common Multiple (LCM) of their individual periods. Next Simultaneous Event = Initial Time + LCM(t_1, t_2, t_3, \dots) Crucially, all time units must be converted to the same metric (e.g., minutes) before calculating the LCM. Step-by-Step Solution: Given: • Device 1: 30 minutes • Device 2: 1 hour = 60 minutes • Device 3: 1(1/2) hour = 90 minutes • Device 4: 1 hour 45 minutes = 105 minutes • Start Time: 12 noon Calculation: 1. Find the LCM of 30, 60, 90, and 105: • 30 = 2 × 3 × 5 • 60 = 2^2 × 3 × 5 • 90 = 2 × 3^2 × 5 • 105 = 3 × 5 × 7 2. Take the highest power of each prime factor: • LCM = 2^2 × 3^2 × 5 × 7 = 4 × 9 × 5 × 7 • LCM = 1260 minutes 3. Convert the LCM back into hours: Hours = (1260/60) = 21 hours 4. Add 21 hours to the starting time (12 noon): • 12 noon + 12 hours = 12 midnight • 12 midnight + 9 hours = 9 a.m. (the next day) Exam Strategy & Shortcut: Instead of full prime factorization, use the step-by-step LCM method: • LCM(30, 60) = 60 • LCM(60, 90) = 180 • Now find LCM(180, 105) • Both are divisible by 15: 180 = 15 × 12 and 105 = 15 × 7. • Since 12 and 7 are co-prime, LCM = 15 × 12 × 7 = 180 × 7 = 1260 minutes. Dividing 1260 by 60 mentally (126 / 6) gives 21 hours instantly. Common Pitfall: Failing to correctly convert mixed time formats (1(1/2) hours or 1 hour 45 minutes) strictly into minutes. Trying to find the LCM using decimal hours (like 1.5 or 1.75) without applying fraction LCM rules will lead to incorrect or messy calculations. Final Answer: Therefore, the correct answer is 9 a.m.

Question 16

Three wheels can complete 40, 24 and 16 revolutions per minute respectively. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

Accepted Answer

Concept & Strategy: To find when multiple repeating cyclical events synchronize again, we must calculate the Least Common Multiple (LCM) of their individual time periods. First, we must convert the given speeds (revolutions per minute) into the time taken for a *single* revolution. Time per revolution = \frac{Total Time}{Number of Revolutions} For fractions, the LCM is calculated as: LCM of fractions = \frac{LCM of Numerators}{HCF of Denominators} Step-by-Step Solution: Given: • Speed of Wheel 1 = 40 rev/min • Speed of Wheel 2 = 24 rev/min • Speed of Wheel 3 = 16 rev/min Calculation: 1. Find the time taken for one complete revolution for each wheel in seconds (since options are primarily in seconds and minutes, seconds is a better base unit): • Wheel 1: 60 seconds / 40 rev = (60/40) = (3/2) seconds • Wheel 2: 60 seconds / 24 rev = (60/24) = (5/2) seconds • Wheel 3: 60 seconds / 16 rev = (60/16) = (15/4) seconds 2. Calculate the LCM of these fractional time periods ((3/2), (5/2), (15/4)): • Find the LCM of the numerators (3, 5, 15): The lowest common multiple of 3, 5, and 15 is 15. • Find the HCF of the denominators (2, 2, 4): The highest common factor of 2, 2, and 4 is 2. 3. Apply the fraction formula: Simultaneous Time = (15/2) seconds 4. Convert to a mixed fraction to match the options: (15/2) seconds = 7.5 seconds = 7(1/2) sec Exam Strategy & Shortcut: Always quickly convert "rate" to "time" before finding the LCM. Taking the LCM of 40, 24, and 16 gives 240, which would incorrectly imply minutes or revolutions. By converting to seconds first ((3/2), (5/2), (15/4)), you immediately see the numerator must be a multiple of 15, guiding you directly to 15/2. Common Pitfall: The most dangerous pitfall is finding the LCM of the given numbers (40, 24, 16) directly. This gives the number of *revolutions* until they align, not the *time*. Always ensure you are finding the LCM of the actual time intervals between the events. Final Answer: Therefore, the correct answer is 7(1/2) sec.

Question 17

A pendulum strikes 5 times in 3 seconds and another pendulum strikes 7 times in 4 seconds. If both pendulums start striking at the same time, how many clear strikes can be listened in 1 minute?

Accepted Answer

Concept & Logic: To find the total number of "clear strikes" heard from two different pendulums, we must use the principle of inclusion-exclusion. A clear strike is heard whenever *either* pendulum strikes. If they strike at the exact same moment, they blend into one clear strike. |A \cup B| = |A| + |B| - |A \cap B| Where |A| and |B| are the number of strikes of each pendulum, and |A \cap B| is the number of simultaneous strikes. Step-by-Step Solution: Given: • Pendulum A rate: 5 strikes in 3 seconds \implies 1 strike every (3/5) seconds. • Pendulum B rate: 7 strikes in 4 seconds \implies 1 strike every (4/7) seconds. • Total time window: 1 minute = 60 seconds. • (Note: Based on standard exam interpretations matching the options, we evaluate the strikes occurring *within* the 60 seconds, excluding the initial simultaneous strike at t=0). Calculation: 1. Calculate individual strikes in 60 seconds: • Pendulum A strikes: (60/3/5) = 60 × (5/3) = 100 strikes • Pendulum B strikes: (60/4/7) = 60 × (7/4) = 105 strikes 2. Calculate the interval of simultaneous strikes: • Find the LCM of the time intervals ((3/5) and (4/7)). • LCM = \frac{LCM(3, 4)}{HCF(5, 7)} = (12/1) = 12 seconds. • They strike together every 12 seconds. 3. Calculate the number of simultaneous strikes in 60 seconds: • Simultaneous strikes = (60/12) = 5 strikes. 4. Calculate the total clear strikes heard: • Total Clear Strikes = 100 + 105 - 5 = 200. Exam Strategy & Shortcut: Instead of writing out fraction formulas, think of it as rates: Pendulum A does (5/3) strikes/sec. In 60s, it does 60 × (5/3) = 100. Pendulum B does (7/4) strikes/sec. In 60s, it does 60 × (7/4) = 105. Total independent strikes = 205. The overlap happens at the LCM of their fractional periods ((3/5) and (4/7)), which is 12 seconds. 60 / 12 = 5 overlaps. Subtract the overlaps: 205 - 5 = 200. Common Pitfall: A major point of confusion is whether to count the strike at t=0. If you calculate (60 / interval) + 1, you get 101, 106, and 6 overlaps, yielding 201. Since 201 is not an option, the question framework clearly asks for the strikes that occur *after* the initial release during the 1-minute span. Always let the options guide the strictness of the boundary conditions. Final Answer: Therefore, the correct answer is 200.

Question 18

Find the HCF of 132, 204 and 228.

Accepted Answer

Concept & Strategy: The Highest Common Factor (HCF) of a set of numbers is the largest number that divides all of them without leaving a remainder. The fastest way to find the HCF of large numbers is using the Difference Method: The HCF of any two numbers must also be a factor of their difference. HCF(a, b) must divide |a - b| Step-by-Step Solution: Given: • Numbers: 132, 204, and 228. Calculation: 1. Find the differences between the adjacent numbers: • Difference 1: 204 - 132 = 72 • Difference 2: 228 - 204 = 24 2. The HCF of the three numbers must be a factor of the smallest difference, which is 24. 3. Look at the given options to find factors of 24: • (a) 12 - Is a factor of 24. • (b) 18 - Not a factor of 24. • (c) 6 - Is a factor of 24. • (d) 21 - Not a factor of 24. 4. Test the highest possible candidate first, which is 12: • 132 \div 12 = 11 (Divisible) • 204 \div 12 = 17 (Divisible) • 228 \div 12 = 19 (Divisible) 5. Since 11, 17, and 19 are prime to each other (co-prime), 12 is indeed the highest common factor. Exam Strategy & Shortcut: Use the difference method combined with option elimination. The difference between 228 and 204 is exactly 24. The HCF *cannot* be larger than 24, and it *must* divide 24 perfectly. Looking at the options (12, 18, 6, 21), 18 and 21 are instantly eliminated. Between 12 and 6, always test the larger number first since you are looking for the *Highest* Common Factor. Common Pitfall: Resorting to full prime factorization for all three numbers (132 = 2^2 × 3 × 11, etc.) takes too much time under exam conditions and increases the chance of minor calculation errors. Use the difference method to save time. Final Answer: Therefore, the correct answer is 12.

Question 19

If three numbers are $2a$, $5a$ and $7a$, what will be their LCM?

Accepted Answer

Concept & Formula: The Least Common Multiple (LCM) of algebraic terms is found by multiplying the LCM of their numerical coefficients by the highest power of each variable present. LCM(cx, cy) = LCM(c) × Highest Power of Variables Step-by-Step Solution: Given: • The three numbers are 2a, 5a, and 7a. Calculation: 1. Separate the numerical coefficients from the variables: • Coefficients: 2, 5, 7 • Variable part: a, a, a 2. Find the LCM of the numerical coefficients: • Since 2, 5, and 7 are all prime numbers, their LCM is simply their product. • LCM(2, 5, 7) = 2 × 5 × 7 = 70 3. Find the LCM of the variable parts: • The highest power of the variable 'a' across all three terms is just a^1 (or a). 4. Combine both parts: • Total LCM = 70 × a = 70a Exam Strategy & Shortcut: Recognize immediately that 2, 5, and 7 share no common factors (they are mutually prime). Therefore, you just multiply them (2 × 5 × 7 = 70). Because the variable 'a' is common to all terms and raised to the first power in each, it simply tags along once. 70 × a = 70a. Common Pitfall: A very common mistake in algebra is multiplying the variables together as if they were independent distinct terms, resulting in 70a^3. Remember, LCM looks for the *highest power* of the common base present in any single term, it does not multiply the bases together. Final Answer: Therefore, the correct answer is 70a.

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